CHAPTER 6comsics.usm.my/tlyoon/teaching/ZCT104_0708/CHAP6.pdf · 2019. 8. 5. · physics with Bethe Printed with FinePrint - purchase at . 13 PYQ 2.7, Final Exam 2003/04 • A large

1

CHAPTER 6CHAPTER 6

Very Brief introduction to Very Brief introduction to

Quantum mechanicsQuantum mechanics

2

A beam of light if pictured as monochromatic wave (λ, ν)

A = 1 unit area

Intensity of the light beam is

λ

A beam of light pictured in terms of photons A = 1

unit

area

Intensity of the light beam is I = Nhν

N = average number of photons per unit time crossing unit area

perpendicular to the direction of propagation

E=hν

2

0 EcI ε=

Intensity = energy crossing one unit area per unit

time. I is in unit of joule per m2 per second

Probabilistic interpretation of matter Probabilistic interpretation of matter

wavewave

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3

•• Consider a beam of lightConsider a beam of light

•• In wave picture, In wave picture, EE = = EE00sin(sin(kxkx––ωωtt), electric ), electric

field in radiationfield in radiation

•• Intensity of radiation in wave picture isIntensity of radiation in wave picture is2

0I cEε=•• On the other hand, in the photon picture, On the other hand, in the photon picture, II = = NhNhνν

•• Correspondence principle: what is explained in the Correspondence principle: what is explained in the

wave picture has to be consistent with what is wave picture has to be consistent with what is

explained in the photon picture in the limit explained in the photon picture in the limit

NN��infinityinfinity: : νε NhEcI == 2

0

Probability of observing a photonProbability of observing a photon

4

Statistical interpretation of radiationStatistical interpretation of radiation

•• The probability of observing a photon at a point in unit The probability of observing a photon at a point in unit time is proportional to time is proportional to NN

•• However, since However, since

•• the probability of observing a photon must also the probability of observing a photon must also

•• This means that the probability of observing a photon at This means that the probability of observing a photon at any point in space is proportional to the square of the any point in space is proportional to the square of the averaged electric field strength at that pointaveraged electric field strength at that point

ProbProb (x) (x)

2E∝2

0 EcNh εν =

2E∝

Square of the mean of the square of

the wave field amplitude


5

What is the physical interpretation of What is the physical interpretation of

matter wave?matter wave?•• we will call the mathematical representation of the de we will call the mathematical representation of the de BroglieBroglie’’ss wave / matter wave / matter

wave associated with a given particle (or an physical entity) aswave associated with a given particle (or an physical entity) as

The wave function, The wave function, ΨΨΨΨΨΨΨΨ

•• We wish to answer the following questions:We wish to answer the following questions:

•• Where is exactly the particle located within Where is exactly the particle located within ∆∆xx? the locality of a particle ? the locality of a particle becomes fuzzy when itbecomes fuzzy when it’’s represented by its matter wave. We can no more tell s represented by its matter wave. We can no more tell for sure where it is exactly located.for sure where it is exactly located.

•• Recall that in the case of conventional wave physics, |field ampRecall that in the case of conventional wave physics, |field amplitudelitude ||||||||2222 2222 is is proportional to the intensity of the wave). Now, what does |proportional to the intensity of the wave). Now, what does |ΨΨΨΨΨΨΨΨ ||||||||22222222 physically physically mean? mean?

6

Probabilistic interpretation of (the Probabilistic interpretation of (the

square of) matter wavesquare of) matter wave

•• As seen in the case of radiation field, As seen in the case of radiation field,

|electric field|electric field’’s amplitudes amplitude||||||||2222 2222 is proportional to the is proportional to the probability of finding a photonprobability of finding a photon

•• In exact analogy to the statistical interpretation of In exact analogy to the statistical interpretation of the radiation field, the radiation field,

•• PP((xx) = ) = ||ΨΨΨΨ ΨΨΨΨ ||||||||22222222 is interpreted as the probability is interpreted as the probability density of observing a material particledensity of observing a material particle

•• More quantitatively, More quantitatively,

•• Probability for a particle to be found between Probability for a particle to be found between point a and b ispoint a and b is

∫∫ Ψ==≤≤b

a

b

a

dxtxdxxPbxap 2|),(|)()(


7

2| ( , ) | is the probability to find the

particle between and

b

ab

a

p x t dx

a b

= Ψ∫

•• It value is given by the area under the curve of It value is given by the area under the curve of

probability density between probability density between a a and and bb

8

Expectation valueExpectation value

•• Any physical observable in quantum mechanics, Any physical observable in quantum mechanics, O O (which is a (which is a function of position, function of position, xx), when measured repeatedly, will yield ), when measured repeatedly, will yield an expectation value of given byan expectation value of given by

•• Example, Example, O O can be the potential energy, position, etc. can be the potential energy, position, etc.

•• (Note: the above statement is not applicable to energy and (Note: the above statement is not applicable to energy and linear momentum because they cannot be express explicitly as linear momentum because they cannot be express explicitly as a function of a function of xx due to uncertainty principle)due to uncertainty principle)……

2*

* *

O dx O dx

O

dx dx

∞ ∞

−∞ −∞∞ ∞

−∞ −∞

Ψ Ψ Ψ

= =

ΨΨ ΨΨ

∫ ∫

∫ ∫


9

Example of expectation value: Example of expectation value:

average position measured for a average position measured for a

quantum particlequantum particle

•• If the position of a quantum particle is If the position of a quantum particle is

measured repeatedly with the same initial measured repeatedly with the same initial

conditions, the averaged value of the conditions, the averaged value of the

position measured is given by position measured is given by

2

2

1

x dx

x x dx

∞

∞

−∞

−∞

Ψ

= = Ψ∫

∫

10

ExampleExample

•• A particle limited to the A particle limited to the xx axis has the wave axis has the wave

function function Ψ Ψ = = axax between between xx=0 and =0 and xx=1; =1; Ψ Ψ = =

0 else where. 0 else where.

•• (a) Find the probability that the particle can (a) Find the probability that the particle can

be found between be found between xx=0.45 and =0.45 and xx=0.55. =0.55.

•• (b) Find the expectation value <(b) Find the expectation value <x> x> of the of the

particleparticle’’s positions position


11

SolutionSolution

•• (a) the probability is(a) the probability is

•• (b) The expectation value is(b) The expectation value is

0.550.55 32 2 2 2

0.45 0.45

0.02513

xdx x dx a a

∞ ∞

−∞

Ψ = = =

∫ ∫

11 3 22 3 2

0 04 4

x ax x dx x dx a

∞

−∞

= Ψ = = =

∫ ∫

12

Max Born and probabilistic Max Born and probabilistic

interpretationinterpretation

•• Hence, a particle’s wave Hence, a particle’s wave function gives rise to a function gives rise to a probabilistic probabilistic interpretationinterpretation of the of the position of a particleposition of a particle

•• Max Born in 1926Max Born in 1926

German-British physicist who worked on the mathematical basis for

quantum mechanics. Born's most important contribution was his

suggestion that the absolute square of the wavefunction in the

Schrödinger equation was a measure of the probability of finding

the particle at a given location. Born shared the 1954 Nobel Prize in

physics with Bethe


13

PYQ 2.7, Final Exam 2003/04PYQ 2.7, Final Exam 2003/04

•• A large value of the probability density of an A large value of the probability density of an

atomic electron at a certain place and time atomic electron at a certain place and time

signifies that the electron signifies that the electron

•• A.A. is likely to be found thereis likely to be found there

•• B.B. is certain to be found thereis certain to be found there

•• C. C. has a great deal of energy therehas a great deal of energy there

•• D.D. has a great deal of chargehas a great deal of charge

•• E. E. is unlikely to be found thereis unlikely to be found there• ANS:A, Modern physical technique, Beiser, MCP 25, pg. 802

14

Particle in in an infinite wellParticle in in an infinite well

(sometimes called particle in a box)(sometimes called particle in a box)

•• Imagine that we put particle Imagine that we put particle

(e.g. an electron) into an (e.g. an electron) into an

““infinite wellinfinite well”” with width with width L L

(e.g. a potential trap with (e.g. a potential trap with

sufficiently high barrier)sufficiently high barrier)

•• In other words, the particle In other words, the particle

is confined within 0 < is confined within 0 < xx < < LL

•• In Newtonian view the In Newtonian view the

particle is traveling along a particle is traveling along a

straight line bouncing straight line bouncing

between two rigid walls between two rigid walls


15

•• The The ‘‘particleparticle’’ is no more pictured as a particle is no more pictured as a particle

bouncing between the walls but a de bouncing between the walls but a de BroglieBroglie wave that wave that

is trapped inside the infinite quantum well, in which is trapped inside the infinite quantum well, in which

they form standing wavesthey form standing waves

However, in quantum view, particle However, in quantum view, particle

becomes wavebecomes wave……

16

Particle forms standing wave Particle forms standing wave

within the infinite wellwithin the infinite well

•• How would the wave How would the wave function of the function of the particle behave inside particle behave inside the well?the well?

•• They form standing They form standing waves which are waves which are confined within confined within

0 0 ≤≤ xx≤≤ LL


17

Standing wave in generalStanding wave in general•• Shown below are standing waves which ends are Shown below are standing waves which ends are

fixed at fixed at xx = 0 and = 0 and xx = = LL

•• For standing wave, the speed is constant, (For standing wave, the speed is constant, (vv = = λ λ ff = =

constant)constant)

18

Mathematical description of standing Mathematical description of standing

waveswaves

•• In general, the equation that describes a standing wave In general, the equation that describes a standing wave (with a constant width (with a constant width LL) is simply: ) is simply:

LL = = nnλλnn/2/2

n n = 1, 2, 3, = 1, 2, 3, …… (positive, discrete integer)(positive, discrete integer)

•• n n characterisescharacterises the the ““modemode”” of the standing waveof the standing wave

•• nn = 1 mode is called the = 1 mode is called the ‘‘fundamentalfundamental’’ or the first or the first harmonicharmonic

•• nn = 2 is called the second harmonics, etc.= 2 is called the second harmonics, etc.

•• λλnnare the wavelengths associated with the are the wavelengths associated with the nn--thth mode mode

standing wavesstanding waves

•• The lengths of The lengths of λλnnis is ““quantisedquantised”” as it can take only as it can take only

discrete values according to discrete values according to λλnn= 2= 2LL//nn


19

Energy of the particle in the boxEnergy of the particle in the box

•• Recall thatRecall that

•• For such a free particle that forms standing waves in the box, iFor such a free particle that forms standing waves in the box, it t

has no potential energy has no potential energy

•• It has all of its mechanical energy in the form of kinetic energIt has all of its mechanical energy in the form of kinetic energy y

only only

•• Hence, for the region 0 < Hence, for the region 0 < xx < < L , L , we write the total energy of we write the total energy of

the particle asthe particle as

<<

≥≤∞=

Lx

LxxxV

0,0

,0,)(

EE = = K K + + VV = = pp22/2/2mm + 0 = + 0 = pp22/2/2mm

20

Energies of the particle are Energies of the particle are

quantisedquantised

•• Due to the quantisation of the standing wave Due to the quantisation of the standing wave

(which comes in the form of (which comes in the form of λλnn = 2= 2LL//nn), the ), the

momentum of the particle must also be quantised momentum of the particle must also be quantised

due to de due to de Broglie’sBroglie’s postulate:postulate:

L

nhhpp

n

n2

==→λ

It follows that the total energy of the particle It follows that the total energy of the particle

is also quantised:is also quantised:2

22

22

22 mLn

m

pEE nn

�π==→


21

2

22

2

2

2

22

282 mLn

mL

hn

m

pE nn

�π===

The The nn = 1 state is a characteristic state called the ground = 1 state is a characteristic state called the ground

state = the state with lowest possible energy (also called state = the state with lowest possible energy (also called

zerozero--point energy )point energy )

Ground state is usually used as the reference state when Ground state is usually used as the reference state when

we refer to ``excited stateswe refer to ``excited states’’’’ ((nn = 2, 3 or higher)= 2, 3 or higher)

The total energy of the The total energy of the nn--thth state can be expressed in term state can be expressed in term

of the ground state energy asof the ground state energy as

((nn = 1,2,3,4= 1,2,3,4……))

The higher The higher nn the larger is the energy levelthe larger is the energy level

2

22

02

)1(mL

EnEn�π

=≡=

0

2EnEn =

22

•• Some terminologySome terminology

•• nn = 1 corresponds to the ground state= 1 corresponds to the ground state

•• nn = 2 corresponds to the first excited state, etc= 2 corresponds to the first excited state, etc

•• Note that lowest possible energy for a Note that lowest possible energy for a particle in the box is not zero but particle in the box is not zero but

EE00 (= (= EE11 ), the zero), the zero--point energy.point energy.

•• This a result consistent with the This a result consistent with the Heisenberg uncertainty principleHeisenberg uncertainty principle

n = 1 is the ground

state (fundamental

mode): 2 nodes, 1

antinode

n = 2 is the first

excited state, 3

nodes, 2 antinodes

n = 3 is the second

excited state, 4

nodes, 3 antinodes


23

Simple analogy

• Cars moving in the right lane on the highway are in

‘excited states’ as they must travel faster (at least

according to the traffic rules). Cars travelling in the left

lane are in the ``ground state’’ as they can move with a

relaxingly lower speed. Cars in the excited states must

finally resume to the ground state (i.e. back to the left

lane) when they slow down

Ground state excited states

24

Example on energy levelsExample on energy levels

•• Consider an electron confined by electrical Consider an electron confined by electrical

force to an infinitely deep potential well whose force to an infinitely deep potential well whose

length length LL is 100 pm, which is roughly one is 100 pm, which is roughly one

atomic diameter. What are the energies of its atomic diameter. What are the energies of its

three lowest allowed states and of the state three lowest allowed states and of the state

with with nn = 15?= 15?

•• SOLUTIONSOLUTION

•• For For nn = 1, the ground state, we have = 1, the ground state, we have

( )( )( )

eV7.37J103.6m10100kg101.9

Js1063.6

8)1( 18

21231

234

2

2

2

1 =×=××

×== −

−−

−

Lm

hE

e


25

eV8481eV7.37225)15(

eV339eV7.379)3(

eV150eV7.374)2(

1

2

15

1

2

3

1

2

2

=×==

=×==

=×==

EE

EE

EE

•• The energy of the remaining states (The energy of the remaining states (nn=2,3,15) =2,3,15)

are are

26

•• When electron makes a transition from the When electron makes a transition from the nn = = 3 excited state back to the ground state, does 3 excited state back to the ground state, does the energy of the system increase or decrease?the energy of the system increase or decrease?

•• Solution:Solution:

•• The energy of the system decreases as energy The energy of the system decreases as energy drops from drops from 339339 eVeV to 150 to 150 eVeV

•• The lost amount |The lost amount |∆∆EE| = | = EE3 3 -- EE11 = = 339339 eVeV –– 150 150 eVeV is radiated away in the form of is radiated away in the form of electromagnetic wave with wavelength electromagnetic wave with wavelength λ λ obeying obeying ∆∆EE = = hchc//λλ

Question continuedQuestion continued


27

n = 1

n = 3

Photon with

λ = xx nmExampleExample

Radiation emitted during deRadiation emitted during de--excitationexcitation

•• Calculate the wavelength of the Calculate the wavelength of the electromagnetic radiation electromagnetic radiation emitted when the excited emitted when the excited system at system at n n = 3 in the previous = 3 in the previous example deexample de--excites to its excites to its ground stateground state

•• SolutionSolution

λ λ == hchc/|/|∆∆E|E|

= 1240 nm. = 1240 nm. eVeV / (|E/ (|E3 3 -- EE11|)|)

= 1240 nm. eV/(= 1240 nm. eV/(339339 eVeV––150 150 eVeV) )

= = xxxx nmnm

28

ExampleExample

A macroscopic particleA macroscopic particle’’s quantum states quantum state

•• Consider a 1 microgram speck of dust moving Consider a 1 microgram speck of dust moving

back and forth between two rigid walls back and forth between two rigid walls

separated by 0.1 mm. It moves so slowly that it separated by 0.1 mm. It moves so slowly that it

takes 100 s for the particle to cross this gap. takes 100 s for the particle to cross this gap.

What quantum number describes this motion?What quantum number describes this motion?


29

SolutionSolution

•• The energy of the particle is The energy of the particle is

( ) ( ) J105s/m101kg1012

1

2

1)( 222692 −−− ×=×××=== mvKE

•• Solving for Solving for nn in in 2

22

2

2mLnEn

�π=

•• yields yields 141038 ×≈= mEh

Ln

•• This is a very large numberThis is a very large number

•• It is experimentally impossible to distinguish between It is experimentally impossible to distinguish between

the n = 3 x 10the n = 3 x 101414 and n = 1 + (3 x 10and n = 1 + (3 x 101414) states, so that ) states, so that

the quantized nature of this motion would never reveal the quantized nature of this motion would never reveal

itself itself

30

•• The quantum states of a macroscopic particle The quantum states of a macroscopic particle

cannot be experimentally discerned (as seen in cannot be experimentally discerned (as seen in

previous example)previous example)

•• Effectively its quantum states appear as a Effectively its quantum states appear as a

continuumcontinuum

E(n=1014) = 5x10-22J

allowed energies in classical system

– appear continuous (such as

energy carried by a wave; total

mechanical energy of an orbiting

planet, etc.)

descret energies in quantised

system – discrete (such as energy

levels in an atom, energies carried

by a photon)

∆E≈ 5x10-22/1014

=1.67x10-36 = 10-17 eV

is too tiny to the discerned


31

PYQ 4(a) Final Exam 2003/04PYQ 4(a) Final Exam 2003/04

•• An electron is contained in a oneAn electron is contained in a one--dimensional dimensional

box of width 0.100 nm. Using the particlebox of width 0.100 nm. Using the particle--inin--aa--

box model,box model,

•• (i) (i) Calculate the Calculate the nn = 1 energy level and = 1 energy level and n n = 4 = 4

energy level for the electron in energy level for the electron in eVeV..

•• (ii)(ii) Find the wavelength of the photon (in nm) Find the wavelength of the photon (in nm)

in making transitions that will eventually get it in making transitions that will eventually get it

from the from the thethe nn = 4 to = 4 to nn = 1 state = 1 state

• Serway solution manual 2, Q33, pg. 380, modified

32

SolutionSolution•• 4a(i) In the particle4a(i) In the particle--inin--aa--box model, standing wave is box model, standing wave is

formed in the box of dimension formed in the box of dimension LL: :

•• The energy of the particle in the box is given by The energy of the particle in the box is given by

•• 4a(ii)4a(ii)

•• The wavelength of the photon going from The wavelength of the photon going from nn = 4 to = 4 to nn = = 1 is 1 is λλ = = hchc/(/(EE66 -- EE11))

•• = 1240 = 1240 eVeV nm/ (603 nm/ (603 –– 37.7) 37.7) eVeV = = 2.2 nm2.2 nm

n

Ln

2=λ

( )2

222

2

2222

282

/

2 Lm

n

Lm

hn

m

h

m

pEK

eee

n

e

n

nn

�πλ=====

2 2

1 237.7 eV

2 e

Em L

π= =

� 2

4 14 603 eVE E= =


33

Example on the probabilistic Example on the probabilistic

interpretation: interpretation:

Where in the well the particle spend Where in the well the particle spend

most of its time?most of its time?

•• The particle spend most of its time in places The particle spend most of its time in places

where its probability to be found is largestwhere its probability to be found is largest

•• Find, for the Find, for the nn = 1 and for = 1 and for nn =3 quantum states =3 quantum states

respectively, the points where the electron is respectively, the points where the electron is

most likely to be foundmost likely to be found

34

SolutionSolution•• For electron in the n = 1 state, For electron in the n = 1 state,

the probability to find the probability to find the the

particle is highest at particle is highest at xx = = LL/2/2

•• Hence electron in the n =1 state Hence electron in the n =1 state

spend most of its time there spend most of its time there

compared to other placescompared to other places

•• For electron in the n = 3 state, the probability to find For electron in the n = 3 state, the probability to find the the

particle is highest at particle is highest at xx = = LL/6,/6,LL/2, 5/2, 5LL/6/6

•• Hence electron in the n =3 state spend most of its time at Hence electron in the n =3 state spend most of its time at

this three placesthis three places


35

Boundary conditions and Boundary conditions and

normalisationnormalisation of the wave function of the wave function

in the infinite wellin the infinite well

•• Due to the probabilistic interpretation of the Due to the probabilistic interpretation of the

wave function, the probability density wave function, the probability density PP((xx) = ) =

||ΨΨ||2 2 must be such that must be such that

•• PP((xx) = |) = |ΨΨ||2 2 > 0 for 0 < > 0 for 0 < xx < < LL

•• The particle has no where to be found at the The particle has no where to be found at the

boundary as well as outside the well, boundary as well as outside the well, i.ei.e PP((xx) = ) =

||ΨΨ||2 2 = 0 for = 0 for x x ≤≤ 0 and x 0 and x ≥≥ LL

36

•• TheThe probability density is probability density is zero at the boundaries zero at the boundaries

•• Inside the well, the particle is Inside the well, the particle is

bouncing back and forth bouncing back and forth

between the wallsbetween the walls

•• It is obvious that it must exist It is obvious that it must exist

within somewhere within the within somewhere within the

wellwell

•• This means:This means:

1||)(0

2 =Ψ= ∫∫∞

∞−

L

dxdxxP


37

•• is called the is called the normalisationnormalisation condition of the wave condition of the wave

functionfunction

•• It represents the physical fact that the particle is It represents the physical fact that the particle is

contained inside the well and the integrated contained inside the well and the integrated

possibility to find it inside the well must be 1possibility to find it inside the well must be 1

•• The The normalisationnormalisation condition will be used to condition will be used to

determine the determine the normalisatonnormalisaton constant when we constant when we

solve for the wave function in the solve for the wave function in the SchrodinderSchrodinder

equationequation

1||)(0

2 =Ψ= ∫∫∞

∞−

L

dxdxxP

38

See if you could answer this See if you could answer this

questionquestion

•• Can you list down the main differences Can you list down the main differences

between the particlebetween the particle--inin--aa--box system box system

(infinite square well) and the Bohr’s (infinite square well) and the Bohr’s

hydrogen like atom? E.g. their energies hydrogen like atom? E.g. their energies

level, their quantum number, their energy level, their quantum number, their energy

gap as a function of gap as a function of nn, the sign of the , the sign of the

energies, the potential etc. energies, the potential etc.


39

SchrodingerSchrodinger EquationEquation

Schrödinger, Erwin (1887-1961),

Austrian physicist and Nobel laureate.

Schrödinger formulated the theory of

wave mechanics, which describes the

behavior of the tiny particles that make

up matter in terms of waves.

Schrödinger formulated the

Schrödinger wave equation to describe

the behavior of electrons (tiny,

negatively charged particles) in atoms.

For this achievement, he was awarded

the 1933 Nobel Prize in physics with

British physicist Paul Dirac

40

What is the general equation that What is the general equation that

governs the evolution and governs the evolution and behaviourbehaviour

of the wave function?of the wave function?

•• Consider a particle subjected to some timeConsider a particle subjected to some time--

independent but spaceindependent but space--dependent potential dependent potential VV((xx) )

within some boundaries within some boundaries

•• The The behaviourbehaviour of a particle subjected to a timeof a particle subjected to a time--

independent potential is governed by the famous (1independent potential is governed by the famous (1--

D, time independent, non relativistic) D, time independent, non relativistic) SchrodingerSchrodinger

equation:equation:

( ) 0)()(

2 2

22

=−+∂

∂xVE

x

x

mψ

ψ�


41

How to derive the T.I.S.E How to derive the T.I.S.E

•• 1) Energy must be conserved: 1) Energy must be conserved: EE = = KK + + UU

•• 2) Must be consistent with de 2) Must be consistent with de BrolieBrolie hypothesis thathypothesis that

p = p = h/h/λλ

•• 3)3) Mathematically wellMathematically well--behaved and sensible (e.g. behaved and sensible (e.g.

finite, single valued, linear so that superposition finite, single valued, linear so that superposition

prevails, conserved in probability etc.)prevails, conserved in probability etc.)

•• Read the Read the mswordmsword notes or text books for more notes or text books for more

technical details (which we will skip here)technical details (which we will skip here)

42

Energy of the particleEnergy of the particle

•• The kinetic energy of a particle subjected to The kinetic energy of a particle subjected to potential potential VV((xx) is ) is

K (= pK (= p22/2m) = E /2m) = E –– V V

•• EE is conserved if there is no net change in the total mechanical is conserved if there is no net change in the total mechanical energy between the particle and the surroundingenergy between the particle and the surrounding

(Recall that this is just the definition of total mechanical ene(Recall that this is just the definition of total mechanical energy) rgy)

•• It is essential to relate the de It is essential to relate the de BroglieBroglie wavelength to the energies of wavelength to the energies of the particle:the particle:

λλ = = hh / / pp = = hh / / √√[2[2mm((EE--VV)])]

•• Note that, as Note that, as VV��0, the above equation reduces to the no0, the above equation reduces to the no--potential potential case (as we have discussed earlier)case (as we have discussed earlier)

λλ = = hh / / pp�� hh / / √√[2[2mEmE], where ], where EE = = KK onlyonly

V(x)E, K

λλ


43

Infinite potential revisitedInfinite potential revisited

•• Armed with the T.I.S.E we now revisit the Armed with the T.I.S.E we now revisit the

particle in the infinite wellparticle in the infinite well

•• By using appropriate boundary condition to the By using appropriate boundary condition to the

T.I.S.E, the solution of T.I.S.E for the wave T.I.S.E, the solution of T.I.S.E for the wave

function function Ψ Ψ should reproduces the quantisation should reproduces the quantisation

of energy level as have been deduced earlier, of energy level as have been deduced earlier,

i.e. i.e. 2

222

2mL

nEn

�π=

In the next slide we will need to do some mathematics to solve for ΨΨ((xx) ) in the second

order differential equation of TISE to recover this result. This is a more formal way

compared to the previous standing waves argument which is more qualitative

44

Why do we need to solve the Why do we need to solve the

ShrodingerShrodinger equation?equation?•• The potential The potential VV((xx) represents the environmental influence on the particle) represents the environmental influence on the particle

•• Knowledge of the solution to the T.I.S.E, i.e. Knowledge of the solution to the T.I.S.E, i.e. ψψ((xx) allows us to obtain ) allows us to obtain essential physical information of the particle (which is subjectessential physical information of the particle (which is subjected to the ed to the influence of the external potential influence of the external potential VV((xx) ), ) ), e.ge.g the probability of its existence the probability of its existence in certain space interval, its momentum, energies etc.in certain space interval, its momentum, energies etc.

Take a classical example: A particle that are subjected to a graTake a classical example: A particle that are subjected to a gravity field vity field UU((xx) ) = = GMmGMm//rr2 2 is governed by the Newton equations of motion, is governed by the Newton equations of motion,

2

2

2 dt

rdm

r

GMm=−

•• Solution of this equation of motion allows us to predict, e.g. tSolution of this equation of motion allows us to predict, e.g. the he position of the object position of the object mm as a function of time, as a function of time, r r = = rr((tt), its ), its instantaneous momentum, energies, etc.instantaneous momentum, energies, etc.


45

S.E. is the quantum equivalent of the S.E. is the quantum equivalent of the

Newton’s law of motionNewton’s law of motion

•• The equivalent of “Newton laws of motion” The equivalent of “Newton laws of motion”

for quantum particles = for quantum particles = ShroedingerShroedinger

equationequation

•• Solving for the wave function in the S.E. Solving for the wave function in the S.E.

allows us to extract all possible physical allows us to extract all possible physical

information about the particle (energy, information about the particle (energy,

expectation values for position, momentum, expectation values for position, momentum,

etc.)etc.)

46

The infinite well in the light of TISEThe infinite well in the light of TISE

<<

≥≤∞=

Lx

LxxxV

0,0

,0,)(

( ) 0)()(

2 2

22

=−+∂

∂xVE

x

x

mψ

ψ�

Plug the potential function V(x)

into the T.I.S.E

Within 0 < x < L, V (x) = 0, hence the

TISE becomes

)()(2)(

2

22

2

xBxEm

x

xψψ

ψ−≡−=

∂

∂

�


47

)()(

2

2

2

xBx

xψ

ψ−=

∂

∂

2

2 2

�

mEB = This term contain the information of the energies of

the particle, which in terns governs the behaviour

(manifested in terms of its mathematical solution) of

ψ (x) inside the well. Note that in a fixed quantum

state n, B is a constant because E is conserved.

However, if the particle jumps to a state n’≠ n, E

takes on other values. In this case, E is not conserved

because there is an net change in the total energy of

the system due to interactions with external

environment (e.g. the particle is excited by external

photon)

The behavior of the particle inside

the box is governed by the equation

If you still recall the elementary mathematics of second order differential

equations, you will recognise that the solution to the above TISE is simply

BxCBxAx cossin)( +=ψWhere A, C are constants to be determined by ultilising the boundary conditions

pertaining to the infinite well system

48

You can prove that indeed You can prove that indeed

BxCBxAx cossin)( +=ψ

is the solution to the TISE is the solution to the TISE )()(

2

2

2

xBx

xψ

ψ−=

∂

∂

•• I will show the steps in the following:I will show the steps in the following:

•• Mathematically, to show that EQ 1 is a solution to EQ Mathematically, to show that EQ 1 is a solution to EQ 2, we just need to show that when EQ1 is plugged into 2, we just need to show that when EQ1 is plugged into the LHS of EQ. 2, the resultant expression is the same the LHS of EQ. 2, the resultant expression is the same as the expression to the RHS of EQ. 2.as the expression to the RHS of EQ. 2.

(EQ 1)

(EQ 2)


49

BxCBxAx cossin)( +=ψ into the LHS of EQ 2:

[ ]

[ ]

[ ]EQ2ofRHS)(

cossin

cossin

sincos

cossin)(

2

2

22

2

2

2

2

=−=

+−=

−−=

−∂

∂=

+∂

∂=

∂

∂

xB

BxCBxAB

BxCBBxAB

BxBCBxBAx

BxCBxAxx

x

ψ

ψ

Plug

Proven that EQ1 is indeed the solution to EQ2

50

Boundary conditionsBoundary conditions•• Next, we would like to solve for the constants Next, we would like to solve for the constants AA,, CC in in

the solution the solution ψψ((xx), as well as the constraint that is ), as well as the constraint that is imposed on the constant imposed on the constant BB

•• We know that the wave function forms nodes at the We know that the wave function forms nodes at the boundaries. Translate this boundary conditions into boundaries. Translate this boundary conditions into mathematical terms, this simply means mathematical terms, this simply means

ψψ((xx = = 0) = 0) = ψψ((xx = L= L) = 0) = 0


51

•• First, First,

•• Plug Plug ψψ((xx = = 0) = 0 into 0) = 0 into

ψ ψ = = AAsinsinBxBx + + CCcoscosBxBx, , we obtainwe obtain

ψ ψ ((x=x=0)0))) = 0 = = 0 = AAsinsin 00 + C + C coscos 00 = C= C

•• i.ei.e, , CC = 0= 0

•• Hence the solution is reduced to Hence the solution is reduced to

ψ ψ ((xx))= = AAsinsinBxBx

52

•• Next we apply the second boundary condition Next we apply the second boundary condition

ψψ ((x = Lx = L) = 0 = ) = 0 = AAsinsin(BL(BL))

•• Only either Only either AA or or sin(sin(BLBL) must be zero but not both) must be zero but not both

•• AA cannot be zero else this would mean cannot be zero else this would mean ψψ ((xx) is zero ) is zero everywhere inside the box, conflicting the fact that the everywhere inside the box, conflicting the fact that the particle must exist inside the boxparticle must exist inside the box

•• The upshot is: The upshot is: AA cannot be zero cannot be zero


53

•• This means it must be This means it must be sinsinBLBL = = 0, or in other words0, or in other words

•• B = n B = n π/ π/ L L ≡≡ BBnn, n = , n = 1,2,3,1,2,3,……

•• nn is used to is used to characterisecharacterise the quantum states of the quantum states of ψψnn ((xx))

•• B is B is characterisedcharacterised by the positive integer by the positive integer nn, hence we use , hence we use BBnn instead of Binstead of B

•• The relationship The relationship BBnn = = nnππ//L L translates into the familiar translates into the familiar quantisationquantisation of energy of energy condition:condition:

•• ((BBnn = n= nππ//L)L)22 ��2

22

2

2

22

2

2

2

2

mLnE

L

nmEB n

nn

�

�

ππ=⇒==

54

�� Hence, up to this stage, the solution is Hence, up to this stage, the solution is

��ψψnn((xx) = ) = AAnnsinsin((nnππxx/L/L), ), nn = 1, 2, 3,= 1, 2, 3,……for 0 < x < Lfor 0 < x < L

��ψψnn((xx) = 0 elsewhere (outside the box)) = 0 elsewhere (outside the box)

��The constant The constant AAnn is yet unknown up to nowis yet unknown up to now

��We can solve for We can solve for AAnn by applying another by applying another ““boundary conditionboundary condition”” –– the the normalisationnormalisationcondition that: condition that:

1)()(0

22 == ∫∫∞

∞−

L

nn dxxdxx ψψ

The area

under the

curves of

|Ψn|2 =1

for every

n


55

Solve for Solve for AAnn with with normalisationnormalisation

•• thusthus

12

)(sin)()(2

0

22

0

22 ==== ∫∫∫∞

∞−

LAdx

L

xnAdxxdxx n

L

n

L

nn

πψψ

LAn

2=

•• We hence arrive at the final solution thatWe hence arrive at the final solution that

��ψψnn((xx) = (2/L)) = (2/L)1/21/2sinsin((nnππx/Lx/L), ), nn = 1, 2, 3,= 1, 2, 3,……for 0 < for 0 < xx < < LL

��ψψnn((xx) = 0 elsewhere (i.e. outside the box)) = 0 elsewhere (i.e. outside the box)

56

ExampleExample•• An electron is trapped in a oneAn electron is trapped in a one--

dimensional region of length dimensional region of length LL = = 1.01.0××1010--1010 m. m.

•• (a) How much energy must be (a) How much energy must be supplied to excite the electron supplied to excite the electron from the ground state to the first from the ground state to the first state? state?

•• (b) In the ground state, what is (b) In the ground state, what is the probability of finding the the probability of finding the electron in the region from electron in the region from

xx = 0.090 = 0.090 ×× 1010--1010 m to 0.110 m to 0.110 ××1010--1010 m? m?

•• (c) In the first excited state, what (c) In the first excited state, what is the probability of finding the is the probability of finding the electron between electron between

xx = 0 and = 0 and x x = 0.250 = 0.250 ×× 1010--1010 m? m? 0.5A 1A0.25A


57

SolutionsSolutions

eV372 2

22

01 ==≡mL

EEπ�(a) eV148)2( 0

2

0

2

2 === EEnE

eV111|| 02 =−=∆⇒ EEE

(b)

0038.02

sin2

1

sin2

)(

11.0

09.0

22

0211

2

1

2

1

2

1

=

−=

==≤≤

=

=

= ∫∫�

�

Ax

Ax

x

x

x

x

n

L

x

L

x

dxL

x

LdxxxxP

π

π

πψ

(c) For n = 2, ;2

sin2

2L

x

L

πψ =

25.04

sin4

1

2sin

2)(

25.0

0

22

2212

2

1

2

1

2

1

=

−=

==≤≤

=

=

= ∫∫�

Ax

x

x

x

x

x

n

L

x

L

x

dxL

x

LdxxxxP

π

π

πψ

On average the particle in

the n = 2 state spend 25% of

its time in the region

between x=0 and x=0.25 A

For ground state

On average the particle in

the ground state spend

only 0.04 % of its time in

the region between

x=0.11A and x=0.09 A

58

The nightmare The nightmare

of a lengthy calculationof a lengthy calculation


59

Quantum tunnelingQuantum tunneling

•• In the infinite quantum well, there are In the infinite quantum well, there are

regions where the particle is regions where the particle is ““forbiddenforbidden”” to to

appearappear V �infinity V �infinity

II

Allowed region

where particle

can be found

I

Forbidden region

where particle

cannot be found

because ψ = 0

everywhere

before x < 0

III

Forbidden region

where particle

cannot be found

because ψ = 0

everywhere after

x > L

ψ(x=0)=0 ψ(x=L)=0n = 1

60

Finite quantum wellFinite quantum well

•• The fact that The fact that yy is 0 everywhere x is 0 everywhere x ≤≤0, x 0, x ≥≥ L L is because of the is because of the infiniteness of the potential, infiniteness of the potential, VV��∞∞

•• If If VV has only finite height, the has only finite height, the solution to the TISE will be solution to the TISE will be modified such that a nonmodified such that a non--zero value zero value of of yy can exist beyond the boundaries can exist beyond the boundaries at at x = x = 0 and 0 and x x = = LL

•• In this case, the pertaining In this case, the pertaining boundaries conditions are boundaries conditions are

)()(),0()0( LxLxxx IIIIIIII ====== ψψψψ

Lx

III

Lx

II

x

II

x

I

dx

d

dx

d

dx

d

dx

d

====

==ψψψψ

,00


61

•• For such finite well, the wave function is not vanished at the bFor such finite well, the wave function is not vanished at the boundaries, and may oundaries, and may extent into the region I, III which is not allowed in the infiniextent into the region I, III which is not allowed in the infinite potential limitte potential limit

•• Such Such ψψ that penetrates beyond the classically forbidden regions diminthat penetrates beyond the classically forbidden regions diminishes very ishes very fast (exponentially) once fast (exponentially) once xx extents beyond x = 0 and extents beyond x = 0 and xx = = LL

•• The mathematical solution for the wave function in the The mathematical solution for the wave function in the ““classically forbiddenclassically forbidden””regions are regions are

••

≥≠−

≤≠=

−

+

LxCxA

xCxAx

,0)exp(

0,0)exp()(ψ

The total energy of the particle

E = K inside the well.

The height of the potential well V is

larger than E for a particle trapped

inside the well

Hence, classically, the particle

inside the well would not have

enough kinetic energy to overcome

the potential barrier and escape

into the forbidden regions I, III

V

E1 = K1

E2 = K2

However, in QM, there is a slight chance to find

the particle outside the well due to the quantum

tunneling effect

62

•• The quantum tunnelling effect allows a The quantum tunnelling effect allows a

confined particle within a finite potential confined particle within a finite potential

well to penetrate through the classically well to penetrate through the classically

impenetrable potential wallimpenetrable potential wall

Hard

and

high

wall,

VE

Hard

and

high

wall,

VE

After many many

times of banging

the wall

Quantum tunneling effect


63

Why tunneling phenomena can Why tunneling phenomena can

happenhappen

•• ItIt’’s due to the continuity requirement of the wave function s due to the continuity requirement of the wave function at the boundaries when solving the T.I.S.Eat the boundaries when solving the T.I.S.E

•• The wave function cannot just The wave function cannot just ““die offdie off”” suddenly at the suddenly at the boundaries of a finite potential wellboundaries of a finite potential well

•• The wave function can only diminishes in an exponential The wave function can only diminishes in an exponential manner which then allow the wave function to extent manner which then allow the wave function to extent slightly beyond the boundariesslightly beyond the boundaries

•• The quantum tunneling effect is a manifestation of the The quantum tunneling effect is a manifestation of the wave nature of particle, which is in turns governed by the wave nature of particle, which is in turns governed by the T.I.S.E. T.I.S.E.

•• In classical physics, particles are just particles, hence never In classical physics, particles are just particles, hence never display such tunneling effectdisplay such tunneling effect

≥≠−

≤≠=

−

+

LxCxA

xCxAx

,0)exp(

0,0)exp()(ψ

64

Quantum tunneling effectQuantum tunneling effect


65

Real example Real example

of tunneling of tunneling

phenomena:phenomena:

alpha decayalpha decay

66

Real example of tunneling phenomena:Real example of tunneling phenomena:

Atomic force microscopeAtomic force microscope


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